1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <vector>
 5 #include <cstdio>
 6 using namespace std;
 7 int dp[25][1000];
 8 //int path[25][1000];
 9 vector<int> path[21][801];
10 int sub[1000];
11 int add[1000];
12 int n,m,fix;
13 int main ()
14 {
15     int num=0;
16     while (cin>>n>>m&&n&&m)  {
17         memset (dp,-1,sizeof(dp));
18         for (int i=0;i<=m;i++)
19             for (int j=0;j<801;j++)
20                 path[i][j].clear();
21         for (int i=1;i<=n;i++) {
22             int d,p; cin>>d>>p;
23             sub[i]=d-p;
24             add[i]=d+p;
25         }
26         fix=20*m;  dp[0][fix]=0;
27         for (int i=1;i<=n;i++) {
28             for (int x=m-1;x>=0;x--)
29                 for (int y=0;y<=2*fix;y++) {//因为sub[i]既可以为正也可以为负，所以状态的转移更新的方向 从x=m-1逆推
30                     if (dp[x][y]>=0&&dp[x+1][y+sub[i]]<dp[x][y]+add[i]) {
31                         dp[x+1][y+sub[i]]=dp[x][y]+add[i];
32                         //path[x+1][y+sub[i]] = i;    //这样定义是不行的  因为取m个   当前最优路径是动态的 中间结果可能被再次更新
33                         path[x+1][y+sub[i]] = path[x][y];
34                         path[x+1][y+sub[i]].push_back(i);
35                     }
36                 }
37         }
38         int ans; int a[25];
39         for (int i=0;i<=fix;i++) if (dp[m][fix-i]>=0||dp[m][fix+i]>=0) {ans=i;break;}
40         ans=dp[m][fix+ans]>dp[m][fix-ans]?ans:-ans;//判断ans的正负
41         int sum1=(dp[m][fix+ans]+ans)/2;
42         int sum2=(dp[m][fix+ans]-ans)/2;
43         int v=fix+ans;
44        /* for (int i=m;i>=1;i--) {
45             a[i]=path[i][v];
46             //cout<<v<<endl;
47             v-=sub[a[i]];
48         }*/
49         printf( "Jury #%d\n", ++num );
50         printf( "Best jury has value %d for prosecution and value %d for defence:\n", sum1,sum2);
51        // for (int i=1;i<=m;i++)
52             //cout<<" "<<a[i];
53         for( int i=0; i < m; i++ )
54             printf( " %d", path[m][fix+ans][i]);
55         cout<<"\n\n";
56     }
57     return 0;
58 }